In the proof of the Expectation of Binomial Distribution, why there is a $k$ in the following equation? I understand the part 
is the probability (PMF) of Binomial Distribution.
But the random variable X should be either 0 or 1. Where is this $k$ coming from?
On
The expectation $E(X)= \int_{a}^{b} X f(x)$ means the expectation of the random variable $X$, when $X$ is continuous variable like in uniform of normal distributions. But in the case of discrete distribution $X==0,1,2,3,...=k$ the expectation value is given as $E(k)=\sum_{k=0} k f(k)$. The Binomial and the Poisson distributions are discrete variable distributions where $X=0,1,2,3,...=k$ These two distributions are given as $$f(k)={n \choose k} p^{k}(1-p)^{n-k}, ~~f(k)=\frac{e^{\lambda} \lambda^k}{k!},$$ respectively. Here $p$ and $\lambda$ are the fixed parameters. So the expectation of the discrete variable $k$ in the binomial case is $$E(k)=\sum_{k=0}^{n} k {n \choose k} p^k(1-p)^k$$
It simply results from the application of the expectation definition, which for a discrete variable ranging from $0$ to $\infty$ is
$${\displaystyle \operatorname {E} [X]=\sum _{i=0}^{\infty }x_{i}\,p_{i}}$$
Each term of the sum is the product of a possible value with the corresponding probability.
The same concept is applied for the binomial distribution, so
$${\displaystyle \operatorname {E} [X]=\sum _{k=0}^{\infty }k \binom nk p^k q^{n-k}}$$
which finally leads to $E(X)=np$.
Regarding your doubt about the fact that the random variable should be either $0$ or $1$: in this case, the variable $X$ for which we are searching the expectation does not refer to the outcome of a single trial, but to the total number of successes in $n$ trials. Thus, to calculate $E(X)$, we have to consider all possible values of $X$ (these are represented by $k$ and range from $0$ to $n$) multiplied by the corresponding probability.